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# Sequences & Series Problem

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Find all pairs of positive integers $$(a,n)$$ such that $$n\ge2$$ and

$$a+(a+1)+(a+2)...+(a+n-1)=100$$.

benjamingu22  May 24, 2017
edited by benjamingu22  Jun 1, 2017
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In the given equation, the total number of "a" that exists is n.

Therefore the equation can be simplified to: $$an + (1+2+...+(n-1))$$

1 + 2 + 3 + ... + n = $$\dfrac{n(n+1)}{2}$$

Therefore the whole thing after the plus sign simplifies to: $$\dfrac{(n-1)(1+(n-1))}{2}=\dfrac{n(n-1)}{2}$$

Original equation becomes $$an+\dfrac{n(n-1)}{2}=100$$

And then further simplify: $$\dfrac{1}{2}n^2+an-\dfrac{1}{2}n-100 = 0$$<--- I am trying to make it a quadratic here.

Take out the common factor n: $$\dfrac{1}{2}n^2+n(a-\dfrac{1}{2})-100 = 0$$

To make it easier to solve, multiply each term by 2.

$$n^2 + n(2a - 1)-200=0$$

It now becomes a quadratic equation with a is a constant and solving for n. Then we can find n in terms of a.

$$n=\dfrac{-(2a-1)\pm\sqrt{(2a-1)^2-(4)(1)(-200)}}{2}=\dfrac{1-2a\pm\sqrt{4a^2-4a+801}}{2}$$

Because n is an integer, $$\sqrt{4a^2-4a+801}$$ must be an integer too, to make n an integer.

That implies that $$4a^2 - 4a + 801$$ is a square number.

Also, $$(4a^2 - 4a + 801)\pmod4\equiv 1$$.

We now think of a square number which mod 4 = 1.

The nearest square number to 801 which mod 4 = 1 is 841, therefore we assume that the quadratic there is = 841 first.

$$4a^2 - 4a + 801 = 841\\ a^2 - a = 10\\ a(a-1)= 10$$

No positive integer "a" satisfies the equation.

The second nearest square number to 801 which mod 4 = 1 is 729,

$$4a^2 - 4a + 801 = 729\\ a^2 - a = -72\\ a(a-1)= -72$$

Still no positive integer "a" satisfies the equation.

We should think of a bigger number.

31^2 = 961 and mod 4 = 1.

$$4a^2 - 4a + 801 = 961\\ a^2 - a = 40\\ a(a-1)= 40$$

Still no positive integer "a" satisfies the equation :(

Try 1089

$$4a^2 - 4a + 801 = 1089\\ a^2 - a = 72\\ a(a-1)= 72\\ a = 9$$

Yay! :D

We back-substitute a = 9 into the quadratic formula to find the corresponding n.

$$n=\dfrac{1-2(9)\pm33}{2}\\ n = 8\text{ or }n=-25(\text{rejected})\\ n = 8$$

We get: a = 9, n = 8 (Finally after a lot of work XD)

MaxWong  May 25, 2017